Jump Game 3, BFS Approach
Standard problem for a BFS: https://leetcode.com/problems/jump-game-iii/
1306. Jump Game III
Medium
Given an array of non-negative integers
arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach to any index with value 0.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [4,2,3,0,3,1,2], start = 5 Output: true Explanation: All possible ways to reach at index 3 with value 0 are: index 5 -> index 4 -> index 1 -> index 3 index 5 -> index 6 -> index 4 -> index 1 -> index 3
Example 2:
Input: arr = [4,2,3,0,3,1,2], start = 0 Output: true Explanation: One possible way to reach at index 3 with value 0 is: index 0 -> index 4 -> index 1 -> index 3
Example 3:
Input: arr = [3,0,2,1,2], start = 2 Output: false Explanation: There is no way to reach at index 1 with value 0.
Constraints:
1 <= arr.length <= 5 * 10^40 <= arr[i] < arr.length0 <= start < arr.length
One way to solve it is via Breadth-First-Search, or BFS. Enqueue the initial position, make sure to mark it as visited. Start the processing. If the solution is found, return. Otherwise emqueue if not visited. Very straightforward; slow, but does the trick. Code is down below, cheers, ACC.
public class Solution
{
public bool CanReach(int[] arr, int start)
{
Queue<int> queueIndex = new Queue<int>();
Hashtable visited = new Hashtable();
visited.Add(start, true);
queueIndex.Enqueue(start);
while (queueIndex.Count > 0)
{
int current = queueIndex.Dequeue();
if (arr[current] == 0) return true;
int[] candidates = { current + arr[current], current - arr[current] };
foreach(int candidate in candidates)
{
if (candidate >= 0 && candidate < arr.Length && !visited.ContainsKey(candidate))
{
visited.Add(candidate, true);
queueIndex.Enqueue(candidate);
}
}
}
return false;
}
}
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